 Solution

The chloride complex has Pt(II), $d^8$, whereas the neutral bis(phosphine) complex has Pt(0), $d^{10}$ in the neutral-ethene assignment. The lower-valent, phosphine-supported Pt(0) centre back-donates more strongly into ethene ${\pi }^{\ast }$, weakening and lengthening its internal C–C bond. This accounts for the longer $1.430\text{A˚}$ distance compared with $1.375\text{A˚}$ in the Pt(II) complex.

Both are formally 16-electron complexes: $8+3(2)+2=16$ for the chloride species and $10+2(2)+2=16$ for the phosphine species. Thus the difference is not simply a difference in total valence count. It reflects the energies and occupations of the metal–ligand molecular orbitals. The increased reduction of the ethene ligand gives the Pt(0) complex more metallacyclopropane character; the two descriptions are bonding limits, not distinct products inferred from the distances alone.